Question

The medians of a right-angled triangled which are drawn from the vertices of the acute angles are $6$ cm and $8$ cm, then the length of the hypotenuse is:

• A. $2\sqrt{5}cm$
• B. $3\sqrt{5}cm$
• C. $4\sqrt{5}cm$
• D. $5\sqrt{5}cm$

Answer 1

Answer: (C)

$4\sqrt{5}cm$

In $△ABC$,

$AE$ and $CD$ are the medians of the triangle.

So, $AE=6$ and $CD=8$

Hence, $BE=EC=\frac{1}{2}BC$ and $AD=DB=\frac{1}{2}AB$

In $△ABE$

$A{E}^2=A{B}^2+B{E}^2$

$\therefore 36=A{B}^2+\frac{1}{4}B{C}^2$ ....(1)

In $△BCD$

$C{D}^2=D{B}^2+B{C}^2$

$\therefore 64=\frac{1}{4}A{B}^2+B{C}^2$ ....(2)

Adding (1) and (2), we get

$100=\frac{5}{4}(A{B}^2+B{C}^2)$

$\therefore A{C}^2=80$

$\therefore AC=4\sqrt{5}$ cm